Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.

Video: Exploring Different Ways of Locating Vertices of Quadratics

Tim Burnham

A comparison of methods for calculating the coordinates of the vertex of a quadratic curve. We use completing the square, reading the values from a graph, finding the midpoint of the roots, using differentiation, and using −𝑏/(2𝑎).

13:37

Video Transcript

There are a number of different ways of locating the vertices of a quadratic. And in this video, we’re gonna look at several different methods side by side. We hope you’ll see the advantages and some of the disadvantages of each and maybe even enjoy exploring a bit of math that you haven’t yet encountered.

Let’s consider the quadratic 𝑦 equals two 𝑥 squared minus two 𝑥 minus twelve. Now we know that the general form of a quadratic equation is 𝑦 equals 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐. So this tells us that the 𝑎-value is two, which is positive, so it’s a happy positive smiley curve, and that means it’s gonna have a minimum point as a vertex. So the vertex, the turning point, will be right at the bottom of the curve down here, at the bottom of that smile.

So let’s now try some different ways to find the coordinates of that minimum point. So first, let’s try completing the square, and some people call this vertex format. So we’ve got 𝑦 equals two 𝑥 squared minus two 𝑥 minus twelve. Now to use the completing the square technique, we need to have just one 𝑥 squared so we need to factor out that two from all of the terms. Now luckily on this particular equation, they’re all multiples of two.

So when we factor out the two, we’ve got two lots of 𝑥 squared minus 𝑥 minus six, so the coefficients there are all whole numbers. So if we remember, completing the square is all about looking at these terms here. Now we need to create some parentheses. I’ve got 𝑥 and 𝑥 because 𝑥 times 𝑥 makes 𝑥 squared. And if we take half of this 𝑥-coefficient as the number term, let’s see what happens.

So the coefficient there is actually one for both of these. So half of one is a half. And remember it’s negative one, so actually it’s gonna be negative a half. So we’ve got 𝑥 minus a half times 𝑥 minus a half. So that’s the square. But when I multiply that out, I’ve got 𝑥 times 𝑥, which is equal to 𝑥 squared; 𝑥 times negative a half, which is negative a half 𝑥; and I’ve got 𝑥 times negative a half, which is minus another half 𝑥; and I’ve got negative a half times negative a half, which is positive a quarter. So the negative a half 𝑥s combine together to just give me negative one 𝑥. So this term is correct; that’s what I was looking for. This term is correct; that’s what I was looking for. But I’ve got this extra plus a quarter on the end so I’m gonna subtract that.

So hopefully we could see that this bit here is the same as this bit here. When I multiply out the brackets or the parentheses, I get 𝑥 squared minus 𝑥 plus a quarter. So if I take away a quarter, I’ve just got 𝑥 squared minus 𝑥, and of course I need this other minus six on the end.

So just tidying that up, I’ve got 𝑥 minus a half all squared inside those big square parentheses and negative a quarter take away another six is minus six and a quarter. So now if we multiply the two back in, we’ve got an expression for 𝑦 which is entirely equivalent to the one that we started off with; it’s exactly the same thing, but it’s just rearranged in the different format, in completing the square or the vertex format.

Now it doesn’t matter what value of 𝑥 you plug into this equation. This bit is always gonna be minus twelve and a half. So that’s a constant term. And this bit here, because it’s something squared — You know the smallest I could possibly make that is zero; zero squared is zero. But anything else squared whether it’s a positive or a negative number, when I multiply it by itself, I’m gonna get a positive result. So this-this first term is always gonna be zero or a positive number.

So if I put a value of 𝑥 equal to a half inside the parentheses, I’m gonna have a half minus a half, which is zero. And that’s gonna be zero squared times two, which is just zero. So the 𝑦-value would be negative twelve and a half. But if I use any other value of 𝑥, then I’m gonna end up with a number which is squared, it’s gonna be a positive number, the value of 𝑦 is gonna be bigger than that.

So 𝑥 equals a half then must be the 𝑥-coordinate of the minimum point on that curve. And of course when 𝑥 is a half and this first term turns out to be zero, the 𝑦- coordinate is gonna be zero take away twelve and a half. So the corresponding 𝑦-coordinate is just negative twelve and a half.

So if we’ve got our quadratic in the completing the square or vertex form. It’s just a matter of reading off what is the 𝑥-coordinate that’s gonna generate this bit to be zero, and this will be your corresponding 𝑦-coordinate. So that becomes quite easy.

So let’s have a look at our second method, reading from the graph. The vertex is at the bottom of the curve there, and we just simply have to read off the 𝑥- coordinate and the 𝑦-coordinate. Now that looks pretty straightforward, doesn’t it? But the only problem with it is that — Well, there’s a limit to the amount of accuracy. It depends on the scale of your graph, how accurately you plot your graph. And in this case, I think it- we could be reasonably sure it was a half, negative twelve and a half. But if those answers weren’t quite as exact as that, maybe the answer was nought point four nine recurring and minus twelve point, you know, four five seven or something like that, we would not have been able to read those coordinates with that level of accuracy. So easy method, but definite limitations in terms of accuracy.

So our third method is to find the 𝑥-coordinate of the midpoint of the roots. So if we find out where that curve cuts the 𝑥-axis, here and here, and then we draw a line of symmetry through the curve, through the midpoint of that so if we find the midpoint of those two and we trace that back to here, we will find out the 𝑥-coordinate of that vertex down at the bottom there.

Now the curve cuts the 𝑥-axis there when the 𝑦-coordinate is equal to zero. So if we put 𝑦 equal to zero in our equation and then we can factor that and solve that, then we can find the 𝑥-coordinates of those two points.

So putting 𝑦 equal to zero, we’ve got zero is equal to two 𝑥 squared minus two 𝑥 minus twelve . And we can factor the quadratic on the right-hand side like this: two 𝑥 plus four times 𝑥 plu- 𝑥 minus three.

And when we’ve done that, we know that we got two things multiplied together to give a zero. So one of them has got to be zero. So either this one is zero or this one is zero. And just solving those simple linear equations, we’ve got 𝑥 is equal to negative two or 𝑥 is equal to three.

So to find the midpoint of those two, we just add them together and divide by two; find the mean of the two points. So that gives us negative two plus three all divided by two, which is equal to a half.

Now having done that, we can plug that half back into the original equation up here to find the corresponding 𝑦-coordinate. So substituting the 𝑥-value, a half, into the equation, we’ve got two times a half squared minus two times a half minus twelve. So that’s two times a quarter minus one minus twelve, which is minus twelve and a half.

So that was that method. Now obviously that method breaks down if-if our curve didn’t actually cut the 𝑦-axis anyway. If we didn’t have any roots of that particular quadratic equation, obviously that-that method wouldn’t have been so easy. Alternatively if there had been exactly one root and the curve had touched the 𝑥-axis in just one place, then that would be the vertex. So we’d already know the 𝑥-coordinate of that vertex. We’d just have to plug that back into the original equation to find the corresponding 𝑦-coordinate. So could be quicker, could be slower, could be difficult, it might not factor so well, there might be horrible numbers, but at least we know that is a method that we can use if we need to.

Now the fourth method, many of you who were watching this video won’t know this but you can use differential calculus. If you can differentiate a function, then it’s a way to calculate the gradient of the tangent of the curve at any point. At the minimum point, at the vertex, the gradient will be zero because the tangent will be on a horizontal line.

So differentiation then is a process that generates a little equation that if we put in an 𝑥-coordinate, so this 𝑥-coordinate here, what it does is it says okay this line here is the tangent to the curve at that point. I’ll work out — you know, you put the 𝑥-value into the equation, it tells you the slope of that line. So if I picked this point here, okay, on the vertex, we can see that the tangent will be completely horizontal. If I pick any other point on that curve, then it’s gonna be either sloping upwards or it’s gonna be sloping downwards like this. But at this vertex here, the gradient is gonna be zero; the slope is gonna to be zero. So if I can work out that equation, put the gradient equal to zero into the equation, I can work out the 𝑥-coordinate of this point that way.

So 𝑦 equals two 𝑥 squared minus two 𝑥 minus twelve. And to work out the gradient function, because we’ve just got a simple polynomial here, we’ve got two lots of two times 𝑥 and then we reduce the power of 𝑥 by one. So that becomes two times two 𝑥. And instead of being to the power of two, it’s to the power of one. And then we’ve got — Really this is 𝑥 to the power of one, so this is one times negative two times 𝑥. And then we reduce that power by one, so that’s gonna be 𝑥 to the power of zero. And anything to the power of zero is just one, so actually we’re losing the 𝑥 from that term there.

And when we’re differentiating, this constant term doesn’t have any effect on our gradient function. So this 𝑥 to the power of zero I’m just gonna rub that out now because 𝑥 to the power of zero as we said is just one. So our gradient function becomes four 𝑥 minus two. And as we said when the tangent has a gradient of zero at that vertex, this gradient function will equal zero so we can just put that equal to zero and we can solve this equation for 𝑥.

So adding two to both sides I’ve got four 𝑥 equals two. And then dividing both sides by four, 𝑥 is equal to two quarters, which obviously simplifies to a half. And to work out the corresponding 𝑦-coordinate, we just need to plug that 𝑥 value back into the original equation again, which we just did in the previous example, so we work out that 𝑦 is equal to negative twelve and a half.

So if you happen to know how to differentiate your function using differentiation, it’s a perfectly good acceptable method of working out where the vertex of your quadratic curve is.

While we’re at the calculus, let’s just consider the general quadratic formula. So 𝑦 equals 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐. Now if we differentiate that with respect to 𝑥, we end up with 𝑑𝑦 by 𝑑𝑥 is equal to two 𝑎𝑥 plus 𝑏. So whatever the 𝑎-value is is in here and whatever the 𝑏-value is is in here. Now if we put that equal to zero to try and find out the general 𝑥-coordinate where the vertex is, where the gradient is equal to zero, we rearrange that equation two 𝑎𝑥 equals negative 𝑏. Taking away 𝑏 from both sides, now we divide both sides by two 𝑎

Look what’s popped out: minus 𝑏 over two 𝑎 is the value of the 𝑥-coordinate. So how we came up with that formula then is differentiating the general form of a quadratic equation. That tells us that for the vertex, where the-the gradient of the tangent to that curve will be zero, we know that the 𝑥-coordinate that corresponds with that will be equal to negative 𝑏 over two 𝑎.

So let’s just quickly summarise those methods then. The first one was completing the square, and that was easy because we just had to read off the 𝑥- and 𝑦-coordinates from the equation that we got. But in the equation we had, we had nice easy numbers to work with. But it could be really quite tricky to get the equation in that format, so that’s potentially the drawback of that one.

The second method was just reading it from the graph. And that’s quite easy, to read coordinates from a graph. But of course it could take a long time to plot the graph, and there’s- there are quite strict limits to the accuracy that you can get using that method.

A third method was to find the midpoint of the roots and then to use that 𝑥-coordinate to plug back into the equation to find the 𝑦-coordinate. Now we got loads of ways of finding the roots: we can use the quadratic formula, we know about systematic trial and improvement, we can read the values off a graph, we can use completing the square, or we can use factoring as we did in our example. And on the downside, we have to plug that 𝑥-value into the original equation to find the corresponding 𝑦-coordinate, and that can take quite a long time. Also there might not be any roots of that particular quadratic equation.

The fourth method then was to use differentiation, and that was easy when you know how to do it. So with differentiation, yeah it’s easy when you know how to do it. But you might have to learn calculus, and you’ve still got to plug the 𝑥-coordinate into the original equation to find the corresponding 𝑦-coordinate.

And lastly, we’ve got the simple minus 𝑏 over two 𝑎 to work out the 𝑥-coordinate. Now that’s pretty simple to use. You just have to plug the values 𝑏 and 𝑎 into minus 𝑏 over two 𝑎 and it tells you what the coor- the 𝑥-coordinate of the vertex is. And on the downside, well you’ve gotta remember that formula, that 𝑥 is equal to minus 𝑏 over two 𝑎, and you’ve still got to plug the 𝑥-coordinate into the original equation to find the corresponding 𝑦-coordinate.

So you’ve got five different methods there, all with their own distinct advantages and disadvantages. It’s worth learning all of them so that you’ve got a choice of which ones to use in which situation. I’ve seen some situations one technique will lend itself better than the other to finding the coordinates of your vertex. Okay, good luck with that!